Mechanics 3 - Further kinematics

[mr bob]

I'm studying M3 as one of the three modules for my further maths alevel, and have just started the first chapter. I'm fine with acceleration as a funtion of time, and integrating to find velocity and displacement, and most of acceleration as a function of displacement, but have come stuck on one of the questions:-

A particle p moves along the positive x axis. Its acceleration is (x + 3)ms^-2 when its displacement from the origin O is x meters. Given that initially, when t = 0, the velocity of P is 3ms^-1 in the direction Ox and x = 0. Obtain:-
a) the speed Vms^-1 of P as a funtction of x
b) x as a function of t

I started by trying (1/2)v^2 = INT(X+3)
therefore getting
v^2 = x^2 + 6x + C
where C = 9
therefore:
v^2 = x^2 + 6x + 9

The book's answer however is different, v = x + 3
I don't know how they got that answer, and as for part b) I'm completely lost as i can't figure out how to bring t into the equation.

Any help would be greatly apprectiated,
Thanks,
Bob

 

[qtp]

how can the velocity = x+3 if the acceleration is x+3 ?

 

[mr bob]

I don't know, that's the books answer. It could be wrong though. I know a few were in M1.

 

[vaishakh]

You can find da/dx.
da/dx*v = da/dx*dx/dt = da/dt.

 

[vaishakh]

Your answer is not different anyway. v^2 = x^2 + 6x + 9 = (x + 3)^2.
Thus v = x + 3. So why should you say that your answer is wrong according to the book? Use the same concept I have shown dor the second question to solve the first question. It is tricky but solve it properly.

 

[mr bob]

Ah of course. Silly mistake there, just wasnt thinking. Thank you.

 

[holulumaster]

Here is the answer:

(a) d/dx(0.5v²) = ( x + 3 )
=> v²/2 = x²/2 + 3x + C
=> v² = x² + 6x + C

when t=0, x=0 and v=3
Therefore, c = 9

v² = x² + 6x + 9
=> v² = (x+3)²
=> v = (x+3)

(b) v = x+3
a = x+3

a = dv/dx * dx/dt
=> (x+3) = 1 * (dx/dt)
Integrating both sides... ( replace "{" as integration sign... (im a noob in this forum))

=> {dx/x+3 = { dt
=> ln|x+3| = t + c .eqn 1

when t=0, x=0

=> c = ln3

therefore... from eqn 1...

=> ln|x+3| = t + ln3
=> ln|x+3| - ln3 = t
=> ln|(x+3)/3| = t
=> (x+3)/3 = exp^t
=> x + 3 = 3exp^t
=> x = 3exp^t - 3